In the figure, RGHN is a trapezium and triangles RNL and LRK are isosceles triangles. PF, PH and FH are straight lines. RN = RL = LK. Find
- ∠p
- ∠q
(a)
∠GRK = 180° - ∠p (Interior angles, RG//PH)
∠LKR = 180° - ∠p (Angles on a straight line)
∠LRK = 180° - ∠p (Isosceles triangle)
58° + 180° - ∠p + 180° - ∠p + ∠q + 15° = 180° (Angles on a straight line, FP)
58° + 180° + 180° + 15° - ∠p - ∠p + ∠q = 180°
433° - 2∠p + ∠q = 180°
2∠p - ∠q = 433° - 180°
2∠p - ∠q = 253°
∠q = 2∠p - 253° --- (1)
∠RLN
= ∠RNL
= 2 x (180° - ∠p)
= 360° - 2∠p (Exterior angle of a triangle)
∠q = 180° - (360° - 2∠p) - (360° - 2∠p)
∠q = 180° - 360° + 2∠p - 360° + 2∠p
∠q = 180° - 360° - 360° + 2∠p + 2∠p
∠q = 4∠p - 540° (Angles sum of triangle)
∠q = 4∠p - 540° --- (2)
(2) = (1)
4∠p - 540° = 2∠p - 253°
4∠p - 2∠p= 540° - 253°
2∠p = 287°
∠p
= 287° ÷ 2
= 143.5°
(b)
From (1)
∠q
= 2∠p - 253°
= 287° - 253°
= 34°
Answer(s): (a) 143.5°; (b) 34°